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March 25, 2012Design and Analysis of Computer Algorithm1

Design and Analysis ofComputer AlgorithmLecture 4-4

Pradondet NilaguptaDepartment of Computer Engineering

This lecture note has been modified from lecture notefor 23250 by Prof. Francis Chin

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March 25, 2012Design and Analysis of Computer Algorithm2

Divide and Conquer(Cont.)

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March 25, 2012Design and Analysis of Computer Algorithm3

MULTIPLICATION OF TWO n-BITNUMBERS For this problem, the basic operations should be

bit operations.

Let xand ybe two n-bit numbers, the traditionalmethod requires O(n2) bit operations, say

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March 25, 2012Design and Analysis of Computer Algorithm4

MULTIPLICATION OF TWO n-BITNUMBERS (cont.) Assume n=2m, we treat x and y as two n-bit strings and

partition them into two halves as x = a * b, y= c * d, i.e.,

concatenation of a, b and c, d. As binary numbers, wecan express

xy = (a2n/2 + b) (c2n/2 + d)

= ac2n + (ad + bc) 2n/2 + bd ............ (*)

The above computation of xy reduces to four

multiplications of (n/2)-bit numbers plus some additionsand shifts (multiplications by powers of 2).

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March 25, 2012Design and Analysis of Computer Algorithm5

Two n-bits Number Analysis Let T(n) be the number of bit-operations required

to multiply two n-bit numbers.

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March 25, 2012Design and Analysis of Computer Algorithm6

Two n-bits Number Analysis (cont.)

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March 25, 2012Design and Analysis of Computer Algorithm7

Two n-bits Number Analysis (cont.)

Thus T(n) depends very much the number ofsubproblems and also the size of thesubproblems.

If the number of subproblems were 1, 3, 4, 8,then the algorithms would be of order n, nlog3, n2,n3 respectively.

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March 25, 2012Design and Analysis of Computer Algorithm8

A better Algorithm So, the multiplication of two n-bit numbers

remains O(n2) if the number of half-sizesubproblems remains 4

In order to reduce the time complexity, weattempt to reduce the number of subproblems

(i.e., multiplications ) by a trick which uses moreadditions and shifts, specifically,by reducing one

multiplication with many additions and shifts.

This trick pays off asymptotically, i.e., when nislarge.

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March 25, 2012Design and Analysis of Computer Algorithm9

A better Algorithm (cont.)u= (a+ b)(c+ d)v= a* cw= b* dz= u-v-w= ad+ bc

So, xycan be written as v2n+ z2 n/2 + w(from Eq (*)).

This approach requires only three multiplications of two(n/2)-bit numbers, plus some additions and shifts, to

multiply two n-bit numbers.

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March 25, 2012Design and Analysis of Computer Algorithm10

Analysis One can use the multiplication routine recursively to

evaluate the products u, v, and w. The additions and shiftsrequire O(n) time. Thus the time complexity of multiplyingtwo n-bit numbers is bounded from above by

where kis a constant reflecting the additions and shifts.

From the above discussion, the solution to the recurrenceis then O(nlog3) = O(n1.59).

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March 25, 2012Design and Analysis of Computer Algorithm11

Example

This divide and conquer technique for multiplication of two n-

bit number can also applied to the multiplication problem of

two n-digit integers and n-degree polynomial.

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March 25, 2012Design and Analysis of Computer Algorithm12

MATRIX MULTIPLICATIONThe matrix multiplication can be performed as follows:

This formulation divides the nx nmatrix into n/2 x n/2matrix, and divide the problem into 8 subproblem of size

n/2. (Note that nis used as the input size even though n2

isthe input size ofthe matrix.) This gives the followingrecurrence,

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March 25, 2012Design and Analysis of Computer Algorithm13

Recurrence Relation

T(n) remains O(n3). However, the matrix multiplication can also be

computed as follows.

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March 25, 2012Design and Analysis of Computer Algorithm14

Strassens Matrix Multiplication

Consequently, 7 multiplications and 18additions/subtractions are needed.

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March 25, 2012Design and Analysis of Computer Algorithm15

Strassens Matrix Multiplication Analysis